Solve x=1+4/3sqrt{13+y2+4y}

$$x=1+ \frac{ 4 }{ 3 } \sqrt{ 13+y2+4y }$$

Solve for y

$$y=\frac{9\left(1-x\right)^{2}}{64}-\frac{y_{2}}{4}-\frac{13}{4}$$

$$\frac{3x-3}{4}\geq 0$$

Steps for Solving Radical Equations

Swap sides so that all variable terms are on the left hand side.

$$1+\frac{4}{3}\sqrt{13+y_{2}+4y}=x$$

Subtract 1 from both sides.

$$\frac{4}{3}\sqrt{13+y_{2}+4y}=x-1$$

Divide both sides of the equation by \frac{4}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{4}{3}\sqrt{4y+y_{2}+13}}{\frac{4}{3}}=\frac{x-1}{\frac{4}{3}}$$

Dividing by \frac{4}{3} undoes the multiplication by \frac{4}{3}.

$$\sqrt{4y+y_{2}+13}=\frac{x-1}{\frac{4}{3}}$$

Divide x-1 by \frac{4}{3} by multiplying x-1 by the reciprocal of \frac{4}{3}.

$$\sqrt{4y+y_{2}+13}=\frac{3x-3}{4}$$

Square both sides of the equation.

$$4y+y_{2}+13=\frac{9\left(x-1\right)^{2}}{16}$$

Subtract 13+y_{2} from both sides of the equation.

$$4y+y_{2}+13-\left(y_{2}+13\right)=\frac{9\left(x-1\right)^{2}}{16}-\left(y_{2}+13\right)$$

Subtracting 13+y_{2} from itself leaves 0.

$$4y=\frac{9\left(x-1\right)^{2}}{16}-\left(y_{2}+13\right)$$

Subtract 13+y_{2} from \frac{9\left(-1+x\right)^{2}}{16}.

$$4y=\frac{9\left(x-1\right)^{2}}{16}-y_{2}-13$$

Divide both sides by 4.

$$\frac{4y}{4}=\frac{\frac{9\left(x-1\right)^{2}}{16}-y_{2}-13}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y=\frac{\frac{9\left(x-1\right)^{2}}{16}-y_{2}-13}{4}$$

Divide \frac{9\left(-1+x\right)^{2}}{16}-13-y_{2} by 4.

$$y=\frac{9\left(x-1\right)^{2}}{64}-\frac{y_{2}}{4}-\frac{13}{4}$$

Solve for x (complex solution)

$$x=\frac{4\sqrt{4y+y_{2}+13}}{3}+1$$

Solve for y (complex solution)

$$y=\frac{9\left(1-x\right)^{2}}{64}-\frac{y_{2}}{4}-\frac{13}{4}$$

$$x=1\text{ or }arg(\frac{3-3x}{4})\geq \pi$$

Solve for x

$$x=\frac{4\sqrt{4y+y_{2}+13}}{3}+1$$

$$y_{2}\geq -4y-13$$